Answer
32
Work Step by Step
$n!=n(n-1)(n-2)\cdot...\cdot 2\cdot 1$
$0!=1$
$\displaystyle \left(\begin{array}{l}
n\\
r
\end{array}\right)=\frac{n!}{r!(n-r)!}$
-------------------------
$\left(\begin{array}{l}
5\\
0
\end{array}\right)+\left(\begin{array}{l}
5\\
1
\end{array}\right)+\left(\begin{array}{l}
5\\
2
\end{array}\right)+\left(\begin{array}{l}
5\\
3
\end{array}\right)+\left(\begin{array}{l}
5\\
4
\end{array}\right)+\left(\begin{array}{l}
5\\
5
\end{array}\right)$
$=\displaystyle \frac{5!}{0!5!}+\frac{5!}{1!4!}+\frac{5!}{2!3!}+\frac{5!}{3!2!}+\frac{5!}{4!1!}+\frac{5!}{5!0!}$
$=1+5+10+10+5+1$
$=32$
